/**
 *Description

Alas! A set of D (1 <= D <= 15) diseases (numbered 1..D) is running through the farm. 
Farmer John would like to milk as many of his N (1 <= N <= 1,000) cows as possible. 
If the milked cows carry more than K (1 <= K <= D) different diseases among them, 
then the milk will be too contaminated and will have to be discarded in its entirety. 
Please help determine the largest number of cows FJ can milk without having to discard the milk.
Input

* Line 1: Three space-separated integers: N, D, and K 

* Lines 2..N+1: Line i+1 describes the diseases of cow i with a list of 1 or more space-separated integers. 
* The first integer, d_i, is the count of cow i's diseases; the next d_i integers enumerate the actual diseases. 
* Of course, the list is empty if d_i is 0.
Output

* Line 1: M, the maximum number of cows which can be milked.
Sample Input

6 3 2
0
1 1
1 2
1 3
2 2 1
2 2 1

Sample Output

5
Hint

OUTPUT DETAILS: 

If FJ milks cows 1, 2, 3, 5, and 6, then the milk will have only two diseases (#1 and #2), which is no greater than K (2). 
 */
/*
 * binary encoding the disease: every bit 0: not carrying that disease, 1 carrying
 * now for each cow encode and get a code
 * for d & k: there are C(d,k) combinations, generate these encodings
 * for each combination (c): find cows (w) that carry no other diseases than in this encoding:  that is w&~c==0
 */
import java.util.Scanner;
public class P2436 {
	static int[] store;
	static int sum=0;
	public static void main(String[] args){
		Scanner cin = new Scanner(System.in);
		int n=cin.nextInt(), d=cin.nextInt(), k=cin.nextInt();
		int i,j,x;
		int cow[] = new int[n];
		int len=1;
		for(i=k+1;i<=d;i++)
			len*=i;
		for(i=2;i<=d-k;i++)
			len/=i;
		store=new int[len];
		
		for(x=1;x<=d-k+1;x++)
			combination(d,k,x,0,0);
		
		for(i=0;i<n;i++){
			j=cin.nextInt();
			while(j>0){
				j--;
				cow[i]=cow[i] | (1 <<(cin.nextInt()-1));
			}
		}
		
		int max=0;
		int sum;
		for(int s:store){
			s=~s;
			sum=0;
			for(int c : cow){
				if((s&c)==0){
					sum++;
				}
			}
			if(sum>max)
				max=sum;
		}
		System.out.println(max);
	}
	
	public static void combination(int d, int k, int point, int current, int result){
		result|=1<<(point-1);
		current++;
		if(current==k){
			store[sum]=result;
			sum++;
			return;
		}
		for(int i=point+1;i<=d+current-k+1;i++){
			combination(d,k,i,current,result);
		}
	}
}
